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3 years ago

3m + 5 = 8 modeled by algebra tiles

Mathematics

2 answers:

larisa86 [58]3 years ago

5 0

So you would subtract 5 to each sides and then divide 3 so m equals 1

Lesechka [4]3 years ago

4 0

3m+5=8
subtract 5 from both sides
3m=3
divide both sides by 3
m=1

I hope I helped! :)

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Find the volume of a right circular cone that has a height of 17.1 in and a base with a

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Answer:

Volume of the right circular cone = 1188.79 cubic inches

Step-by-step explanation:

Height of the cone = 17.1 inch

Circumference of the cone= 16.3 inches

Radius = 16.3/2

= 8.15 inch

Volume of the cone is:

$=\pi *r^2*\frac{h}{3}$

As, $\pi =\frac{22}{7}=3.14$

Volume of the cone

$=3.14*(8.15*8.15)*\frac{17.1}{3} \\\\=3.14*(8.15*8.15)*5.7\\\\=3.14*66.42*5.7\\\\=3.14*378.59\\\\=1188.79inch^3$

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3 years ago

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3 years ago

Whats the answer to number 9?

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3 years ago

Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

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Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

= $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\$

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3 years ago